Find the imaginary part of this sum 
Let $$S = e^{i\alpha} + \frac{e^{i3\alpha}}{3} +
 \frac{e^{i5\alpha}}{3^2} + \cdots$$
Find Im$(S)$ and show that it is equal to the sum
$$I = \sin(\alpha) + \frac{\sin(3\alpha)}{3} + \frac{\sin(5\alpha)}{3^2} + \cdots$$

So, I found that $S = \frac{3(3e^{i\alpha} - e^{-i\alpha})}{10 - 6\cos(2\alpha)}$ using the formula for geometric series.
I have a provided answer of $\frac{6\sin(\alpha)}{5 - 3\cos(2\alpha)}$ which I can see that I get if I just take the $\sin(\alpha)$ terms out of the $e$ terms in my numerator; however, why don't I have to change the $\cos(2\alpha)$ term in the denominator? Isn't this part of Re$(S)$? I was confused about his part and was trying to change this term before I looked at the answer.
 A: $$
S = e^{i\alpha} + \frac{e^{i3\alpha}}{3} + \frac{e^{i5\alpha}}{3^2} + \cdots = e^{i\alpha}\left(1 + \frac{e^{2i\alpha}}{3} + \frac{(e^{2i\alpha})^2}{3^2}+ \cdots\right).
$$
The thing inside the parentheses is a geometric series with common ratio $e^{2i\alpha}/3$.  Therefore the expression above is equal to
$$
\frac{e^{i\alpha}}{1-(e^{2i\alpha}/3)} = \frac{3e^{i\alpha}}{3-e^{2i\alpha}}.
$$
This is
$$
\frac{3e^{i\alpha}(3-e^{-2i\alpha})}{(3-e^{2i\alpha})(3-e^{-2i\alpha})} = \frac{9e^{i\alpha}- 3e^{-i\alpha}}{9 - 3e^{2i\alpha} - 3e^{-2i\alpha} + 1} = \frac{6\cos\alpha + 12i\sin\alpha}{10 - 6\cos(2\alpha)}.
$$
That the imaginary part of this is equal to $i$ times that other sum follows from two facts: (1) the terms of the other sum are the imaginary parts of the terms of the first sum; and (2) the imaginary part of the sum equals the sum of the imaginary parts.
A: $$
\begin{align}
S
&=e^{ia}\sum_{k=0}^\infty\left(\frac{e^{i2\alpha}}{3}\right)^k\\
&=\frac{3e^{i\alpha}}{3-e^{i2\alpha}}\cdot\frac{3-e^{-i2\alpha}}{3-e^{-i2\alpha}}\\
&=\frac{9e^{i\alpha}-3e^{-i\alpha}}{10-6\cos(2\alpha)}\\
&=\frac{6\cos(\alpha)+12i\sin(\alpha)}{10-6\cos(2\alpha)}\\
&=\color{red}{\frac{3\cos(\alpha)}{5-3\cos(2\alpha)}}+i\color{green}{\frac{6\sin(\alpha)}{5-3\cos(2\alpha)}}
\end{align}
$$
Therefore,
$$
\mathrm{Im}(S)=\color{green}{\frac{6\sin(\alpha)}{5-3\cos(2\alpha)}}
$$
Since the imaginary part of $\color{blue}{\dfrac{e^{i(2k+1)\alpha}}{3^k}}$ is $\color{orange}{\dfrac{\sin((2k+1)\alpha)}{3^k}}$ we get your second sum
$$
\begin{align}
\mathrm{Im}(S)
&=\mathrm{Im}\left(\sum_{k=0}^\infty\color{blue}{\frac{e^{i(2k+1)\alpha}}{3^k}}\right)\\
&=\sum_{k=0}^\infty\color{orange}{\frac{\sin((2k+1)\alpha)}{3^k}}
\end{align}
$$
